使用PHP显示MySQL临时表的结果-缓存？

Question

I am currently having an issue where MySQL is only displaying 1 of my 3 rows in a dynamic Temporary Table I've created in a PHP page. I can confirm how many rows the TmpTable has via:

$numrows = mysqli_num_rows($doResults);

(returns 3)

But when I do my while ($rows=mysqli_fetch_array($doResults)) { } , only 1 of the 3 rows are returned/displayed. The one row is returned correctly with all fields requested.

Has anyone had any issues with some sort of Caching, etc ... ?

Answer 1

mysqli_fetch_array() returns the next record in the result set, only one record. The name of your variable $row suggests that you expect otherwise.

Try something like

error_reporting(E_ALL); // only for debugging
ini_set('display_errors', 1); // only for debugging

$numrows = mysqli_num_rows($doResults);
echo '<pre>debug: $numrows=', $numrows, "</pre>\n";
$dbgcRow = 0;
while($row=mysqli_fetch_array($doResults)) {
  // row counter / number of the current record
  echo '<pre>debug: $dbgcRow=', ++$dbgcRow, "</pre>\n";
  // print the values of the current record
  echo htmlentities(join(',', $row)), "<br />\n";
}